Using the Left-Inverse to "Solve" an Impossible System of Equations
If $Av=b$, then $TAv=Tb$. The converse is in general not true, as you have noticed. In fact, $Av=b\iff Av-b=0$, but
$$TAv=Tb\iff T(Av-b)=0\iff Av-b\in \ker T$$
Hence, the converse may fail to hold if $\ker T \neq \{0\}$, that is, if $T$ is not injective.
Multiplying both sides by a non invertible matrix can change the solution set. For instance, you get infinitely many solutions when multiplying both sides of any linear system by the zero matrix.
A different case, but similar under several respects, is when you have $x^2=-1$; if you multiply both sides by $x$, you get the equation $x^3=-x$, which has $0$ as a solution. Or, if you square both sides, you get $x^4=1$, which has $1$ and $-1$ has roots. On the other hand, the original equation has no solution (in the real numbers). The main point is that if you do reversible transformations to your equation, you are guaranteed that the solution set doesn't change; with nonreversible transformations, you don't lose solutions, but may add some.
However there is a real-world application of this “wrong” method. If you choose appropriately the left inverse, namely use the Moore-Penrose pseudoinverse, which in this case is $$ A^+=(A^TA)^{-1}A^T $$ where $A$ is the coefficient matrix of the given system, produces a system which features the “least squares solution”.
Since $$ \begin{bmatrix} 4 & 0 & 0 \\ 0 & 5 & 0 \end{bmatrix} \begin{bmatrix} 4 & 0 \\ 0 & 5 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 16 & 0 \\ 0 & 25 \end{bmatrix} $$ we have $$ A^+=\begin{bmatrix} 1/16 & 0 \\ 0 & 1/25 \end{bmatrix} \begin{bmatrix} 4 & 0 & 0 \\ 0 & 5 & 0 \end{bmatrix} = \begin{bmatrix} 1/4 & 0 & 0 \\ 0 & 1/5 & 0 \end{bmatrix} $$ and $$ A^+\begin{bmatrix}1\\1\\1\end{bmatrix}=\begin{bmatrix} 1/4 \\ 1/5 \end{bmatrix} $$
While the given system has no solution, there are reasons why $x_1=1/4$ and $x_2=1/5$ can be considered “almost a solution”.
All you have shown is that if $\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$ is a solution then $\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$ has the form given. This doesn't mean that there really is a solution.